The lifespans of gorillas in a particular zoo are normally distributed. The average gorilla lives $19$ years; the standard deviation is $1.5$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a gorilla living less than $16$ years.
Solution: $19$ $17.5$ $20.5$ $16$ $22$ $14.5$ $23.5$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $19$ years. We know the standard deviation is $1.5$ years, so one standard deviation below the mean is $17.5$ years and one standard deviation above the mean is $20.5$ years. Two standard deviations below the mean is $16$ years and two standard deviations above the mean is $22$ years. Three standard deviations below the mean is $14.5$ years and three standard deviations above the mean is $23.5$ years. We are interested in the probability of a gorilla living less than $16$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the gorillas will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the gorillas will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $16$ years and the other half $({2.5\%})$ will live longer than $22$ years. The probability of a particular gorilla living less than $16$ years is ${2.5\%}$.